Question

tan( a + b) is equal to P and tan(a-b) equal to Q then find tan 2a

Answer 1

Answer:

$\frac{P + Q }{1-PQ}$ is the required value of tan2a .

Step-by-step explanation:

Explanation:

Given , tan(a + b) = P and tan(a - b) = Q

Formula of tan(x+ y ) = $\frac{tanx+ tan y }{1- tanxtany}$

Step 1:

From the given question we have tan2a

And tan 2a can be written as ,

tan2a = tan[(a + b) +(a -b)]

From the formula of tan (x+ y) = $\frac{tanx+ tan y }{1- tanxtany}$

⇒ tan 2a = $\frac{tan(a+b)+ tan (a - b) }{1- tan(a+ b)tan(a- b)}$

Now , put the value of tan(a+ b) and tan(a - b) we get ,

tan2a = $\frac{P + Q }{1-PQ}$

Final answer:

Hence , the value of tan2a is  $\frac{P + Q }{1-PQ}$ .

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Answer 2

Answer:

The value of   $tan2A =\frac{P+Q}{1-PQ}$  .

Step-by-step explanation:

We know that

•   $tan2A = tan[(A+B)+(A-B)]$

•   $tan(x+y) = \frac{tanx+tany}{1-tanx\cdot tany}$

         

Given that   $tan(A+B) = P$   and   $tan(A-B) = Q$

$\implies tan2A = tan[(A+B)+(A-B)]$

                  $= \frac{tan(A+B)+tan(A-B)}{1-tan(A+B)tan(A-B)}$

                  $=\frac{P+Q}{1-PQ}$

Hence,   $tan2A =\frac{P+Q}{1-PQ}$

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